C++ How to combine two signed 8 Bit numbers to a 16 Bit short? Unexplainable results

Your lsb in this case contains 0xfff3. When you OR it with 1 << 8 nothing changes because there is already a 1 in that bit position.

Try short combined = (msb << 8 ) | (lsb & 0xff);

Or using a union:

#include <iostream>

union Combine
{
    short target;
    char dest[ sizeof( short ) ];
};

int main()
{
    Combine cc;
    cc.dest[0] = -13, cc.dest[1] = 1;
    std::cout << cc.target << std::endl;
}

It is possible that lsb is being automatically sign-extended to 16 bits. I notice you only have a problem when it is negative and msb is positive, and that is what you would expect to happen given the way you're using the or operator. Although, you're clearly doing something very strange here. What are you actually trying to do here?